How do you solve $2 {(x + 4)}^{\frac{2}{3}} = 8$ $2(x+4)^(2/3) = 8$ ?

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How do you solve $2 {(x + 4)}^{\frac{2}{3}} = 8$ $2(x+4)^(2/3) = 8$ ?

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Answer 1 · 2015-04-09T17:33:11

Note that $q^{\frac{2}{3}} = {(q^{2})}^{\frac{1}{3}}$ $q^(2/3) = (q^2)^(1/3)$

So
$2 {(x + 4)}^{\frac{2}{3}} = 8$ $2(x+4)^(2/3) = 8$

can be evaluated as
$2 {(x^{2} + 8 x + 16)}^{\frac{1}{3}} = 8$ $2(x^2+8x+16)^(1/3) = 8$

cube both sides giving:
$8 (x^{2} + 8 x + 16) = 8 \times 8 \times 8$ $8 (x^2+8x+16) = 8xx8xx8$
or
$x^{2} + 8 x + 16 = 64$ $x^2+8x+16 = 64$
$\to$ $rarr$
$x^{2} + 8 x - 48 = 0$ $x^2+8x-48=0$

factoring:
$(x + 12) (x - 4) = 0$ $(x+12)(x-4) = 0$

So
$x = - 12$ $x=-12$
or
$x = 4$ $x=4$

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How do you solve $2 {(x + 4)}^{\frac{2}{3}} = 8$ $2(x+4)^(2/3) = 8$ ?

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